Đáp án:
a) \(\sqrt y - \sqrt x \)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;y \ge 0;x \ne y\\
A = \left[ {\dfrac{{x + \sqrt {xy} + y - \sqrt {xy} }}{{\sqrt x + \sqrt y }}} \right]:\left[ {\dfrac{x}{{\sqrt y \left( {\sqrt x + \sqrt y } \right)}} + \dfrac{y}{{\sqrt x \left( {\sqrt y - \sqrt x } \right)}} - \dfrac{{x + y}}{{\sqrt {xy} }}} \right]\\
= \dfrac{{x + y}}{{\sqrt x + \sqrt y }}:\dfrac{{x\left( {\sqrt {xy} - x} \right) + y\left( {\sqrt {xy} + y} \right) - \left( {x + y} \right)\left( {y - x} \right)}}{{\sqrt {xy} \left( {\sqrt x + \sqrt y } \right)\left( {\sqrt y - \sqrt x } \right)}}\\
= \dfrac{{x + y}}{{\sqrt x + \sqrt y }}.\dfrac{{\sqrt {xy} \left( {\sqrt x + \sqrt y } \right)\left( {\sqrt y - \sqrt x } \right)}}{{x\sqrt {xy} - {x^2} + y\sqrt {xy} + {y^2} - {y^2} + {x^2}}}\\
= \dfrac{{x + y}}{{\sqrt x + \sqrt y }}.\dfrac{{\sqrt {xy} \left( {\sqrt x + \sqrt y } \right)\left( {\sqrt y - \sqrt x } \right)}}{{x\sqrt {xy} + y\sqrt {xy} }}\\
= \dfrac{{x + y}}{{\sqrt x + \sqrt y }}.\dfrac{{\sqrt {xy} \left( {\sqrt x + \sqrt y } \right)\left( {\sqrt y - \sqrt x } \right)}}{{\sqrt {xy} \left( {x + y} \right)}}\\
= \sqrt y - \sqrt x
\end{array}\)
